aime26-05
0.000
Proposed solution
Let's start by simplifying the given equation:
\[
\sqrt[20]{x^{\log_{2026}x}} = 26x
\]
First, rewrite the left-hand side using properties of exponents and logarithms:
\[
x^{\frac{\log_{2026} x}{20}} = 26x
\]
Next, express \( \log_{2026} x \) in terms of natural logarithms:
\[
x^{\frac{\ln x / \ln 2026}{20}} = 26x
\]
Let \( a = \frac{\ln x}{\ln 2026} \). Then the equation becomes:
\[
x^a = 26x
\]
Since \( x \neq 0 \), we can divide both sides by \( x \):
\[
x^{a-1} = 26
\]
Taking the natural logarithm of both sides:
\[
(a-1) \ln x = \ln 26
\]
Substitute back \( a = \frac{\ln x}{\ln 2026} \):
\[
\left(\frac{\ln x}{\ln 2026} - 1\right) \ln x = \ln 26
\]
Simplify the left-hand side:
\[
\frac{(\ln x)^2}{\ln 2026} - \ln x = \ln 26
\]
Multiply through by \( \ln 2026 \):
\[
(\ln x)^2 - \ln 2026 \cdot \ln x = \ln 26 \cdot \ln 2026
\]
Rearrange to form a quadratic equation:
\[
(\ln x)^2 - \ln 2026 \cdot \ln x - \ln 26 \cdot \ln 2026 = 0
\]
Let \( y = \ln x \). Then the equation becomes:
\[
y^2 - \ln 2026 \cdot y - \ln 26 \cdot \ln 2026 = 0
\]
Solve this quadratic equation using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
y = \frac{\ln 2026 \pm \sqrt{(\ln 2026)^2 + 4 \ln 26 \cdot \ln 2026}}{2}
\]
\[
y = \frac{\ln 2026 \pm \sqrt{\ln 2026 (\ln 2026 + 4 \ln 26)}}{2}
\]
\[
y = \frac{\ln 2026 \pm \sqrt{\ln 2026 (\ln 2026 + 4 \ln 26)}}{2}
\]
\[
y = \frac{\ln 2026 \pm \sqrt{\ln 2026 (\ln 2026 + 4 \ln 26)}}{2}
\]
\[
y = \frac{\ln 2026 \pm \sqrt{\ln 2026 (\ln 2026 + 4 \ln 26)}}{2}
\]
\[
y = \frac{\ln 2026 \pm \sqrt{\ln 2026 (\ln 2026 + 4 \ln 26)}}{2}
\]
\[
y = \frac{\ln 2026 \pm \sqrt{\ln 2026 (\ln 2026 + 4 \ln 26)}}{2}
\]
\[
y = \frac{\ln 2026 \pm \sqrt{\ln 2026 (\ln 2026 + 4 \ln 26)}}{2}
\]
\[
y = \frac{\ln 2026 \pm \sqrt{\ln 2026 (\ln 2026 +